Finding the Line Integral

**Paymemoney** · May 27th 2011, 04:04 PM

Hi
I am having trouble on working out the following question:

Given C is the anticlockwise circular path $x^2+y^2=4$ , starting and ending at (0,2) evaluate the line integrals $\oint_C F.dr$ where $F = xyi+xyj$

This is what i have done

$r=xi + yj$
$dr = i + j$

$x=2cos(\theta)$ , $y=2sin(\theta)$

$F = (4cos(\theta) sin(\theta))i + (4cos\theta sin\theta )j$

$\int_{0}^(2\pi) (4cos\theta sin\theta)i + (4cos\theta sin\theta )j \cdot i + j$

P.S

**Joanna** · May 27th 2011, 04:34 PM

$dr\neq \hat{\imath} + \hat{\jmath}$
$dr = dx\hat{\imath} + dy \hat{\jmath}$

If
$x=2\cos\theta$ , $y=2\sin\theta$

what are dx and dy?

**Paymemoney** · June 2nd 2011, 11:32 PM

ok i have attempted this question again and this is what i have done but this is incorrect.

This is what i have have done:
$\int_0^{2\pi} 2cos(\theta)+2cos(\theta)2sin(\theta)*2cos(\theta)-2sin(\theta) d(\theta)$

$\int_0^{2\pi} 8cos(\theta)^2 * sin(\theta) - 8cos(\theta) * sin(\theta)^2 + 8cos(\theta)^2 * sin(\theta) - 8cos(\theta)* sin(\theta)^2$

$\int_0^{2\pi} 8(cos(\theta)^2 * sin(\theta) - cos(\theta) * sin(\theta)^2 + cos(\theta)^2 * sin(\theta) - cos(\theta)* sin(\theta)^2)$

**HallsofIvy** · June 3rd 2011, 03:05 AM

Without parentheses, it is almost impossible to read what you have written. With $x= 2 cos(\theta)$ , $y= 2 sin(\theta)$ , $dx= -2 sin(\theta)dt$ and $dy= 2 cos(\theta)dt$ . F(x,y) becomes $4sin(\theta)cos(\theta)i+ 4 sin(\theta)cos(\theta)j$ and so the integral is
$8\int_{\pi/2}^{5\pi/2} (-sin(\theta)cos^2(\theta)+ sin^2(\theta)cos(\theta))dt$
Personally, I would have used $x= 2sin(\theta)$ , $y= 2cos(\theta)$ so I could integrate from 0 to $2\pi$ .

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